Download DIAGONALIZATION OF MATRICES OF CONTINUOUS FUNCTIONS

DIAGONALIZATION OF MATRICES OF CONTINUOUS
FUNCTIONS
EFTON PARK
Definition 1. A (complex) algebra is a ring (A, +, ·) along with a scalar multiplication · : C × A −→ A that makes A into a vector space and such that
λ · (AB) = (λ · A)B = A(λ · B) for all λ in C and A and B in A.
Throughout these notes, we will assume that our algebras A possesses a unit; i.e.,
a multiplicative identity.
Example 1. For each natural number n, the set M(n, C) of n × n matrices is a
ring under matrix addition and multiplication. Scalar multiplication of a matrix A
by a complex number λ is defined by multiplying each entry of A by λ.
Example 2. Let X be a compact Hausdorff space. Let C(X) denote the set of
continuous complex valued functions on X. Pointwise addition, multiplication, and
scalar multiplication make C(X) into an algebra.
Note that if X consists of a single point, C(X) is (isomorphic to) C.
We can combine Examples 1 and 2:
Example 3. For each compact Hausdorff space X and each natural number n, the
set M(n, C(X)) of n × n matrices with entries in C(X) forms an algebra.
Definition 2. Let A and B be elements of an algebra A. We say that A and B
are similar if there exists an invertible element S of A such that B = SAS −1 .
It is easy to show that similarity is an equivalence relation.
Definition 3. A ∗-algebra is an algebra A equipped with an involution; that is, a
map ∗ : A −→ A with the property that (A∗ )∗ = A for all A in A.
In Example 1, the involution is complex conjugate transpose. In Example 2, the
involution is pointwise complex conjugation; i.e., f ∗ (x) := f (x) for every x in X.
The involution in Example 4 is a combination of the involutions in Examples 2 and
3:


∗ 
f11 f21 f31 · · · fn1
f11 f12 f13 · · · f1n
 f21 f22 f23 · · · f2n 
 f12 f22 f32 · · · fn2 




 f31 f32 f33 · · · f3n 



 =  f13 f23 f33 · · · fn3 
 ..



..
..
.
.
.
.
.
.
..
..
.. 
..
..
.. 
 .
 ..
.
.
.
fn1
fn2
fn3
···
fnn
f1n
f2n
f3n
Date: February 28 and March 6, 2012.
2010 Mathematics Subject Classification. 46L05, 47B15, 47A05.
Key words and phrases. unitary equivalence, C ∗ -algebras.
1
···
fnn
2
EFTON PARK
Definition 4. Let A be a ∗-algebra. An element U in A is called unitary if U
is invertible and U −1 = U ∗ . Elements A and B in A are unitarily equivalent if
B = U ∗ AU for some unitary U in A.
Unitary equivalence is an equivalence relation, and for a ∗-algebra, unitary equivalence implies similarity. However, as we shall see in a minute, the converse is not
generally true.
Problem. Determine the unitary equivalence classes of M(n, C(X)). More specifically, determine a complete set of unitary invariants for M(n, C(X)); i.e., a collection of quantities
f1 (A), f2 (A), . . . , fk (A)
for each element A of M(n, C(X)) that have the feature that A and B in M(n, C(X))
are unitarily equivalent if and only if fi (A) = fi (B) for all 1 ≤ i ≤ k.
A reasonable solution to this problem is hopeless in this generality, even if X is a
point. Consider elements in M(n, C) that have the form


1 1 a13 a14 a15 · · · a1n
0 2 1 a24 a25 · · · a2n 


0 0 3
1 a35 · · · a3n 

.
 .. ..
..
..
..
.. 
.
.
. .
.
.
.
.
. 
0
0
0
0
0
···
n
By the Jordan Canonical Form Theorem, each of these matrices is similar to the
diagonal matrix diag(1, 2, 3, . . . , n). However, when n > 2, two such matrices are
unitarily equivalent if and only if they are equal. Thus there are similarity classes
in M(n, C) that are the union of uncountably many unitary equivalences classes!
The situation is somewhat simpler if we restrict our attention to normal matrices.
Definition 5. An element A in a ∗-algebra is normal if AA∗ = A∗ A.
There are many normal elements in a ∗-algebra; given A in A, the elements <(A) :=
1
1
∗
∗
2 (A + A ) and =(A) := 2i (A − A ) are self-adjoint, and therefore normal. Note
∗
that (i=(A)) = −i=(A), so i=(A) is normal; we call such an element skew-adjoint.
Therefore every element of A is the sum of two normal elements (in fact, a selfadjoint element and a skew-adjoint one). But of course A itself may or may not be
normal.
Proposition 4 (Berberian). Two normal elements of M(n, C(X)) (in fact, in any
C ∗ -algebra) are unitarily equivalent if and only if they are similar.
Theorem 5 (Spectral Theorem). Every normal matrix in M(n, C) is unitarily
equivalent to a diagonal matrix.
In this case, the unitary invariants are the roots of the characteristic polynomial
P (λ) := det(λI − A).
Suppose we have A is in M(n, C) and that U ∗ AU = D for some unitary matrix
U and some diagonal matrix D. For each 1 ≤ i ≤ n, let ei denote the vector in
Cn that is 1 in the ith slot and 0 elsewhere. Then for each i, we have Dei = λi
for some complex number λi , and so AU ei = U Dei = U (λi )ei = λi U ei . In other
words, each U ei is an eigenvector for A, and the set {U ei | 1 ≤ i ≤ n} is a complete
set of eigenvectors for A. Therefore the columns of U are the eigenvectors of A.
DIAGONALIZATION OF MATRICES OF CONTINUOUS FUNCTIONS
3
Furthermore, unitary matrices preserve length, so the length of each column of U
equals 1. We shall use these observations shortly.
Question. Let X be a compact Hausdorff space. Is every normal element of
M(n, C(X)) unitarily equivalent to a diagonal matrix?
Example 6. Define A ∈ M(n, C([−1, 1])) as follows:
x x
x
A(x) =
, x ≥ 0,
A(x) =
x x
0
0
,
0
x < 0.
Suppose there exists a unitary matrix U and a diagonal matrix D in M(n, C([−1, 1]))
such that U ∗ AU = D. Then U ∗ (x)A(x)U (x) = D(x) for every −1 ≤ x ≤ 1. For
x > 0,
λ−x
−x
det(λI − A(x)) = det
= (λ − x)2 − x2 = λ2 − 2λx = λ(λ − 2x),
−x
λ−x
f (x)
g(x)
which has eigenvectors of the form
and
for the eigenvalues 0 and
−f (x)
g(x)
2x respectively. Thus for x ≥ 0, we must have one of the following forms:
1
f (x) g(x)
g(x) f (x)
U (x) =
or U (x) =
, |f (x)| = |g(x)| = √ .
−f (x) g(x)
g(x) −f (x)
2
However, for x < 0, we obtain
h(x)
0
0
h(x)
U (x) =
or U (x) =
, |h(x)| = |k(x)| = 1.
0
k(x)
k(x)
0
Clearly, there is no way to choose f , g, h, and k to make U continuous. Thus, while
A(x) is diagonalizable for every x in [−1, 1], the matrix A itself is not diagonalizable.
In Example 6, notice that the dimension of the eigenspace for 0 jumped at x = 0;
this is what caused the problem.
Definition 6. An element A in M(n, C(X)) is multiplicity-free if A(x) has n
distinct roots for each x in X. Equivalently, A is multiplicity-free if the polynomial
P (x, λ) := det(λI − A(x)) has n distinct roots for each x in X.
Is repeated multiplicity the only obstruction to diagonalization?
Definition 7. A simple Weierstrass polynomial of degree n over X is a function
P : X × C −→ C of the form
P (x, λ) = λn + an−1 (x)λn−1 + an−2 λn−2 + · · · + a1 (x)λ + a0 (x),
where a0 , a1 , . . . an−1 are in C(X) and such that for each x, P (x, λ) has n distinct
roots.
Note that if A is multiplicity-free, its characteristic polynomial is a simple Weierstrass polynomial.
Let P be a simple Weierstrass polynomial over X, and let
E := {(x, λ) ∈ X × C | P (x, λ) = 0}.
Projection onto the first coordinate defines a map p : E −→ X. This map is a
covering map, and E is an n-fold cover of X; such a cover is called a polynomial
covering of X.
4
EFTON PARK
Suppose that A in M(n, C(X)) is diagonalizable; choose a unitary U so that U ∗ AU
is diagonal. As we observed earlier, each column of U (x) is an eigenvector for A(x).
For each 1 ≤ i ≤ n, let di (x) be the eigenvalue of A(x) associated to the ith column
of U (x). Because U is a continuous function of X, the functions di : X −→ C are
also continuous. As a consequence, we obtain a continuous global factorization of
the characteristic polynomial:
n
Y
P (x, λ) =
(λ − di (x)).
i=1
1
Example 7. Let X = S and consider the matrix
0 x
A(x) =
.
1 0
The characteristic polynomial is
λ
P (x, λ) = det(λI − A(x)) = det
−1
−x
λ
= λ2 − x.
As we all know from taking complex analysis, this polynomial does not continuously
factor over S 1 . Put another way, P (x, λ) defines a nontrivial double cover of S 1 .
Therefore, even though A is normal and multiplicity-free, it does not diagonalize.
If A in M(n, C(X)) is multiplicity-free and its characteristic polynomial globally
factors into linear factors, is A diagonalizable?
Definition 8. A (complex) vector bundle over X is a topological space V and a
continuous surjection p : V −→ X such that
• p−1 (x) is a complex vector space;
• for each x in X, there exists a neighborhood U of X with the property that
p−1 (U ) is homeomorphic to U × Cn for some natural number n.
If V is homeomorphic to X × Cn for some natural number n, we say that V is a
trivial vector bundle.
Definition 9. An element A of M(n, C(X)) is called a projection if A = A∗ = A2 .
If A is a projection matrix over X, then the range of A at each point x in X
determines a vector bundle ran A over X. Furthermore, up to vector bundle isomorphism, every vector bundle arises in this way.
Example 8. Take X = S 2 , written in cylindrical coordinates (z, θ), and define
√
1
θ 1 − z2
√1 + z
.
A(z, θ) =
1−z
2 θ̄ 1 − z 2
Direct computation show that A is a projection and that its characteristic polynomial
is λ2 − λ = λ(λ − 1). This polynomial is certainly globally factorable into linear
factors. However, the vector bundle ran A is known to be nontrivial. Interesting side
fact: the unit vectors in the range of A at each point determine a topological space
homeomorphic to S 3 , and via this homeomorphism, we obtain a famous continuous
function p : S 3 −→ S 2 called the Hopf vibration.
DIAGONALIZATION OF MATRICES OF CONTINUOUS FUNCTIONS
5
Thus there are at least three obstructions to diagonalizability of matrices over a
compact Hausdorff space X:
• Multiplicities of eigenvalues;
• Existence of nontrivial polynomial covering spaces over X;
• Existence of nontrivial complex vector bundles over X.
It turns out that these are the only obstructions.
If we restrict our attention to multiplicity-free normal matrices, there is a nice result
which tells us about diagonalizability in many cases.
Theorem 9 (Grove-Pedersen). Suppose X is a 2-connected compact CW -complex
and that A ∈ M(n, C(X)) is multiplicity-free. Then A is diagonalizable over X.
Sketch of proof. The hypothesis π1 (X) = 0 implies that the characteristic polynomial P of A globally splits into linear factors, so P defines a trivial polynomial cover
of X, and we can choose continuously varying eigenspaces Qi of A. Because we
are assuming that A is multiplicity-free, these eigenspaces Qi are one-dimensional
and therefore each determine a line bundle ran Qi over X. A line bundle L over
X is completely determined by its first Chern class c1 (L) ∈ H 2 (X, Z). But π1 (X)
and π2 (X) are both trivial by hypothesis, so H 2 (X, Z) = 0 by Hurewicz’s theorem.
Thus we can pick a continuous non-vanishing section fi of ran Qi for each i. By
applying the Gram-Schmidt procedure (which is continuous as a function of X),
we may assume that {f1 , f2 , . . . , fn } are a continuously varying orthonormal frame;
the change of basis matrix U ∈ M(n, C(X)) conjugates A to a diagonal matrix. What if A is not multiplicity-free?
Definition 10. Let X be a topological space. We say that X is σ-compact if it is
the union of a countable number of compact sets. We say that X is sub-Stonean if
any two disjoint open σ-compact subsets of X have disjoint closures.
Example 10. If X is a countable set with the discrete topology, then X is subStonean.
Example 11. Let X be a locally compact σ-compact Hausdorff space and let β(X)
denote the Stone-Cěch compactification of X; roughly speaking, β(X) is the largest
compact topological space in which X is dense. The topological space K(X) :=
β(X) − X is called the corona of X, and with these hypotheses on X, the space
K(X) is sub-Stonean.
Example 12. No infinite first-countable compact Hausdorff space is sub-Stonean.
Theorem 13 (Grove-Pedersen). Suppose X is a compact Hausdorff space with the
property that every self-adjoint element of M(n, C(X)) can be diagonalized. Then
X is sub-Stonean.
Proof. Use Urysohn’s lemma, along with some cleverness.
6
EFTON PARK
Theorem 14. Let X be a compact Hausdoff space. The following statements are
equivalent:
• Every normal matrix over X can be diagonalized.
• The topological space X satisfies the following conditions:
X is sub-Stonean;
the (topological) dimension of X is at most two;
every finite cover of every closed subset of X is trivial;
every complex line bundle over every closed subset of X is trivial.
Example 15. Let Y be a locally compact graph. Then every normal matrix over
the corona of Y can be diagonalized if and only if for some compact subset C of
Y , the complement Y − C is a countable forest; i.e., a countable disjoint union of
trees.
Department of Mathematics, Box 298900, Texas Christian University, Fort Worth, TX
76129
E-mail address: [email protected]